Problem: As a particle moves along the number line, its position at time $t$ is $s(t)$, its velocity is $v(t)$, and its acceleration is $a(t)= 12t-4$. If $v(2) = 4$ and $s(0) = 1$, what is $s(1)$ ? $s(1)=~$
Answer: The antiderivative of $~a(t)~$ is $~v(t)=6t^2-4t+C\,$. We know that $~v(2) = 4\,$, so $~4=6\cdot2^2-4\cdot2+C=16+C\,$. That gives us $~C=-12\,$. Therefore, $~v(t) = 6t^2-4t-12\,$. The antiderivative of $~v(t)~$ is $~s(t)=2t^3-2t^2-12t+K\,$. We know that $~s(0) = 1\,$, so $~K=1\,$. Therefore $~s(t)=2t^3-2t^2-12t+1\,$. Now we can evaluate $~s(1)\,$. $ s(1)=2-2-12+1=-11$